3.1055 \(\int \frac {A+B x}{\sqrt {e x} \sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=300 \[ \frac {\sqrt [4]{a} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \left (\frac {A \sqrt {c}}{\sqrt {a}}+B\right ) \sqrt {\frac {a+b x+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{c^{3/4} \sqrt {e x} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt [4]{a} B \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+b x+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{c^{3/4} \sqrt {e x} \sqrt {a+b x+c x^2}}+\frac {2 B x \sqrt {a+b x+c x^2}}{\sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )} \]

[Out]

2*B*x*(c*x^2+b*x+a)^(1/2)/c^(1/2)/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)-2*a^(1/4)*B*(cos(2*arctan(c^(1/4)*x^(1/2)/a^
(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/a^(1/4))),1/2*(
2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x*c^(1/2))*x^(1/2)*((c*x^2+b*x+a)/(a^(1/2)+x*c^(1/2))^2)^(1/2)/c^(3/4)/(e
*x)^(1/2)/(c*x^2+b*x+a)^(1/2)+a^(1/4)*(cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^
(1/2)/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x*c
^(1/2))*(B+A*c^(1/2)/a^(1/2))*x^(1/2)*((c*x^2+b*x+a)/(a^(1/2)+x*c^(1/2))^2)^(1/2)/c^(3/4)/(e*x)^(1/2)/(c*x^2+b
*x+a)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {841, 839, 1197, 1103, 1195} \[ \frac {\sqrt [4]{a} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \left (\frac {A \sqrt {c}}{\sqrt {a}}+B\right ) \sqrt {\frac {a+b x+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{c^{3/4} \sqrt {e x} \sqrt {a+b x+c x^2}}-\frac {2 \sqrt [4]{a} B \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+b x+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{c^{3/4} \sqrt {e x} \sqrt {a+b x+c x^2}}+\frac {2 B x \sqrt {a+b x+c x^2}}{\sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(Sqrt[e*x]*Sqrt[a + b*x + c*x^2]),x]

[Out]

(2*B*x*Sqrt[a + b*x + c*x^2])/(Sqrt[c]*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (2*a^(1/4)*B*Sqrt[x]*(Sqrt[a] + Sqrt
[c]*x)*Sqrt[(a + b*x + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], (2 - b/(
Sqrt[a]*Sqrt[c]))/4])/(c^(3/4)*Sqrt[e*x]*Sqrt[a + b*x + c*x^2]) + (a^(1/4)*(B + (A*Sqrt[c])/Sqrt[a])*Sqrt[x]*(
Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + b*x + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1
/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(c^(3/4)*Sqrt[e*x]*Sqrt[a + b*x + c*x^2])

Rule 839

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2, Subst[Int[(f +
g*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, b, c, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 841

Int[((f_) + (g_.)*(x_))/(Sqrt[(e_)*(x_)]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[x]/Sq
rt[e*x], Int[(f + g*x)/(Sqrt[x]*Sqrt[a + b*x + c*x^2]), x], x] /; FreeQ[{a, b, c, e, f, g}, x] && NeQ[b^2 - 4*
a*c, 0]

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1197

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]
/; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {A+B x}{\sqrt {e x} \sqrt {a+b x+c x^2}} \, dx &=\frac {\sqrt {x} \int \frac {A+B x}{\sqrt {x} \sqrt {a+b x+c x^2}} \, dx}{\sqrt {e x}}\\ &=\frac {\left (2 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {A+B x^2}{\sqrt {a+b x^2+c x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt {e x}}\\ &=\frac {\left (2 \left (A+\frac {\sqrt {a} B}{\sqrt {c}}\right ) \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2+c x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt {e x}}-\frac {\left (2 \sqrt {a} B \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+b x^2+c x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt {c} \sqrt {e x}}\\ &=\frac {2 B x \sqrt {a+b x+c x^2}}{\sqrt {c} \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {2 \sqrt [4]{a} B \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+b x+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{c^{3/4} \sqrt {e x} \sqrt {a+b x+c x^2}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+b x+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{\sqrt [4]{a} c^{3/4} \sqrt {e x} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 1.70, size = 444, normalized size = 1.48 \[ -\frac {x^2 \left (-\frac {i \sqrt {\frac {4 a}{x \left (\sqrt {b^2-4 a c}+b\right )}+2} \sqrt {\frac {-x \sqrt {b^2-4 a c}+2 a+b x}{b x-x \sqrt {b^2-4 a c}}} \left (B \sqrt {b^2-4 a c}+2 A c-b B\right ) F\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {a}{b+\sqrt {b^2-4 a c}}}}{\sqrt {x}}\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {x}}-\frac {4 B \sqrt {\frac {a}{\sqrt {b^2-4 a c}+b}} (a+x (b+c x))}{x^2}+\frac {i B \left (\sqrt {b^2-4 a c}-b\right ) \sqrt {\frac {4 a}{x \left (\sqrt {b^2-4 a c}+b\right )}+2} \sqrt {\frac {-x \sqrt {b^2-4 a c}+2 a+b x}{b x-x \sqrt {b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {a}{b+\sqrt {b^2-4 a c}}}}{\sqrt {x}}\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {x}}\right )}{2 c \sqrt {e x} \sqrt {\frac {a}{\sqrt {b^2-4 a c}+b}} \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(Sqrt[e*x]*Sqrt[a + b*x + c*x^2]),x]

[Out]

-1/2*(x^2*((-4*B*Sqrt[a/(b + Sqrt[b^2 - 4*a*c])]*(a + x*(b + c*x)))/x^2 + (I*B*(-b + Sqrt[b^2 - 4*a*c])*Sqrt[2
+ (4*a)/((b + Sqrt[b^2 - 4*a*c])*x)]*Sqrt[(2*a + b*x - Sqrt[b^2 - 4*a*c]*x)/(b*x - Sqrt[b^2 - 4*a*c]*x)]*Elli
pticE[I*ArcSinh[(Sqrt[2]*Sqrt[a/(b + Sqrt[b^2 - 4*a*c])])/Sqrt[x]], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*
a*c])])/Sqrt[x] - (I*(-(b*B) + 2*A*c + B*Sqrt[b^2 - 4*a*c])*Sqrt[2 + (4*a)/((b + Sqrt[b^2 - 4*a*c])*x)]*Sqrt[(
2*a + b*x - Sqrt[b^2 - 4*a*c]*x)/(b*x - Sqrt[b^2 - 4*a*c]*x)]*EllipticF[I*ArcSinh[(Sqrt[2]*Sqrt[a/(b + Sqrt[b^
2 - 4*a*c])])/Sqrt[x]], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/Sqrt[x]))/(c*Sqrt[a/(b + Sqrt[b^2 -
4*a*c])]*Sqrt[e*x]*Sqrt[a + x*(b + c*x)])

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} {\left (B x + A\right )} \sqrt {e x}}{c e x^{3} + b e x^{2} + a e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)^(1/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x + a)*(B*x + A)*sqrt(e*x)/(c*e*x^3 + b*e*x^2 + a*e*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x + A}{\sqrt {c x^{2} + b x + a} \sqrt {e x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)^(1/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x + A)/(sqrt(c*x^2 + b*x + a)*sqrt(e*x)), x)

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maple [A]  time = 0.12, size = 538, normalized size = 1.79 \[ \frac {\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {c x}{b +\sqrt {-4 a c +b^{2}}}}\, \left (A b c \EllipticF \left (\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}}{2}\right )+4 B a c \EllipticE \left (\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}}{2}\right )-2 B a c \EllipticF \left (\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}}{2}\right )-B \,b^{2} \EllipticE \left (\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}}{2}\right )+\sqrt {-4 a c +b^{2}}\, A c \EllipticF \left (\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}}{2}\right )-\sqrt {-4 a c +b^{2}}\, B b \EllipticE \left (\sqrt {\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}}{2}\right )\right )}{\sqrt {c \,x^{2}+b x +a}\, \sqrt {e x}\, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x)^(1/2)/(c*x^2+b*x+a)^(1/2),x)

[Out]

1/(c*x^2+b*x+a)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b+(-4*a*c+b^2)^(1/2)))^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2
))/(-4*a*c+b^2)^(1/2))^(1/2)*(-1/(b+(-4*a*c+b^2)^(1/2))*c*x)^(1/2)*(A*EllipticF(((2*c*x+b+(-4*a*c+b^2)^(1/2))/
(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*c*(-4*a*c+b^2)^(1
/2)+A*EllipticF(((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2)
)/(-4*a*c+b^2)^(1/2))^(1/2))*c*b-2*B*EllipticF(((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2
*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*a*c-B*EllipticE(((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b+(
-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*(-4*a*c+b^2)^(1/2)*b+
4*B*EllipticE(((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/
(-4*a*c+b^2)^(1/2))^(1/2))*a*c-B*EllipticE(((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b+(-4*a*c+b^2)^(1/2)))^(1/2),1/2*2^(
1/2)*((b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2))^(1/2))*b^2)/(e*x)^(1/2)/c^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x + A}{\sqrt {c x^{2} + b x + a} \sqrt {e x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)^(1/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/(sqrt(c*x^2 + b*x + a)*sqrt(e*x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x}{\sqrt {e\,x}\,\sqrt {c\,x^2+b\,x+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((e*x)^(1/2)*(a + b*x + c*x^2)^(1/2)),x)

[Out]

int((A + B*x)/((e*x)^(1/2)*(a + b*x + c*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x}{\sqrt {e x} \sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)**(1/2)/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((A + B*x)/(sqrt(e*x)*sqrt(a + b*x + c*x**2)), x)

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